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I suppose we are looking for integer solutions.  since we have irrational numbers, x must be a multiple of 2.  Let us try  0, 2, 4, 6 etc.  But on RHS we have  x/2 in power.  So we need x to be a multiple of 4 to have an integer or rational number on RHS.

let     x = 4

(√2 )⁴ + (√3)⁴ = 2² + 3² = 13
  (√13)^⁴/² = (√13)² = 13
so x = 4  is good.

For values of x > 4 like 8, 12,... , the value on the RHS is much more than the LHS.  So there is only one solution.

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